Cos complexity?

$e^{ix}=\text{cos}x+i\text{sin}x$

Put, $x=i$

$e^{i^2}=\text{cos}(i)+i\text{sin}(i)$

$e^{-1}=\text{cos}(i)+i\text{sin}(i)$ $\rightarrow$ 1

Put, $x=-i$

$e^{-i^2}=\text{cos}(-i)+\text{sin}(-i)$

Since, cosine is an even function, $\text{cos}(-i)=\text{cos}i$

$e=\text{cos}(i)-i\text{sin}(i)$ $\rightarrow$ 2

Adding 1 and 2 , and dividing by 2,

$\large \ \boxed{ cos(i)=\dfrac{e^{-1}+e}{2}≈1.54308}$

Using the following identity of $\cos z$ , where $z$ is a complex number, we have:

$\begin{aligned} \cos \color{#3D99F6}{z} & = \frac{e^{i\color{#3D99F6}{z}}+e^{-i\color{#3D99F6}{z}}}{2} \quad \quad \small \color{#3D99F6}{\text{Putting }z=i} \\ \implies \cos \color{#3D99F6}{i} & = \frac{e^{\color{#3D99F6}{i^2}}+e^{-\color{#3D99F6}{i^2}}}{2} = \frac{e^{-1}+e^1}{2} \approx \boxed{1.543} \end{aligned}$

cosh 1=1.543

$cos(i) = 1+\dfrac{1}{2!}+\dfrac{1}{4!}+\dfrac{1}{6!}+\dfrac{1}{8!}\cdots$ $≈$ 1.543 Very nice observation. Complex cos has crossed its limit.

It's just simply the equivalent of $cosh(1)$